Question

HIEE!! HELP NEEDED

$\frac{cos \: 45}{sec \: 30 \: + \: cosec \: 30 }$

Answer 1

To Find:

The Value of:

$\frac{cos45°}{sec30° + cosec30°}$

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As we know that,

The values of:

$\cos(45°) = \frac{1}{ \sqrt{2} }$

$\sec(30°) = \frac{1}{ \cos(30°) } = \frac{2}{ \sqrt{3} }$

$cosec(30°) = \frac{1}{ \sin(30°) } = 2$

(FROM THE TRIGNOMETRIC TABLE)

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NOW,

ON Substituting the values above into the given expression:

$\frac{cos45°}{sec30° + cosec30°} = \frac{ \frac{1}{ \sqrt{2} } }{ \frac{2}{ \sqrt{3} } + 2}$

$( \frac{1}{ \sqrt{2} } )\div ( \frac{2 +2 \sqrt{3} }{ \sqrt{3} } )$

On rationalising the denominator

of (1/√2)

$( \frac{1}{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } ) \times ( \frac{ \sqrt{3} }{2 + 2 \sqrt{3} } )$

$\frac{ \sqrt{6} }{4} \times \frac{ {1} }{ \sqrt{3} + 1 }$

On rationalising the denominator

of [1/(√3+1)]

$\frac{ \sqrt{6} }{4} \times ( \frac{1(\sqrt{3} - 1 )}{( \sqrt{3} + 1)( \sqrt{3} - 1) } )$

$\frac{ \sqrt{6} }{4} \times \frac{\sqrt{3} - 1}{2}$

$\frac{ \sqrt{18} - \sqrt{6} }{8}$

$\frac{ \sqrt{3 \times 3 \times 2} - \sqrt{6} }{8}$

$= \frac{ 3\sqrt{2} - \sqrt{6} }{8}$

Therefore;

Answer:

$\frac{cos45°}{sec30° + cosec30°} = \frac{ 3\sqrt{2} - \sqrt{6} }{8}$

Answer 2

Question:

find the value of

$\dfrac{ \cos(45°) }{ \sec(30°) + \csc(30°) }$

Used formulas:

cos45° = 1/√2

sec30° = 2/√3

cosec30° = 2

Answer:

$= \dfrac{ \frac{1}{ \sqrt{2} } }{ \frac{2}{ \sqrt{3} } + 2 } \\ \\ = \dfrac{ \frac{1}{ \sqrt{2} } }{ \frac{2 + 2 \sqrt{3} }{ \sqrt{3} } } \\ \\ = \dfrac{ \sqrt{3} }{(2 + 2 \sqrt{3} ) \sqrt{2} } \\ \\ = \dfrac{ \sqrt{3} }{2 \sqrt{2} + 2 \sqrt{6} } \\ \\ = \frac{ \sqrt{3} }{2 \sqrt{2} (1 + \sqrt{3} )}$

Used concepts:

→ trigonometric ratios

→ trigonometric ratios for standard angles